Question

In PHP, I am able to use a normal function as a variable without problem, but I haven’t figured out how to use a static method. Am I just missing the right syntax, or is this not possible?

(EDIT: the first suggested answer does not seem to work. I’ve extended my example to show the errors returned.)

function foo1($a,$b) { return $a/$b; }  class Bar {     static function foo2($a,$b) { return $a/$b; }      public function UseReferences()     {         // WORKS FINE:         $fn = foo1;         print $fn(1,1);          // WORKS FINE:         print self::foo2(2,1);         print Bar::foo2(3,1);          // DOES NOT WORK ... error: Undefined class constant 'foo2'         //$fn = self::foo2;         //print $fn(4,1);          // DOES NOT WORK ... error: Call to undefined function self::foo2()         //$fn = 'self::foo2';         //print $fn(5,1);          // DOES NOT WORK ... error: Call to undefined function Bar::foo2()                 //$fn = 'Bar::foo2';         //print $fn(5,1);       } }  $x = new Bar(); $x->UseReferences(); 

(I am using PHP v5.2.6 — does the answer change depending on version too?)

Answer 1

PHP handles callbacks as strings, not function pointers. The reason your first test works is because the PHP interpreter assumes foo1 as a string. If you have E_NOTICE level error enabled, you should see proof of that.

‘Use of undefined constant foo1 – assumed ‘foo1”

You can’t call static methods this way, unfortunately. The scope (class) is relevant so you need to use call_user_func instead.

<?php  function foo1($a,$b) { return $a/$b; }  class Bar {     public static function foo2($a,$b) { return $a/$b; }      public function UseReferences()     {         $fn = 'foo1';         echo $fn(6,3);          $fn = array( 'self', 'foo2' );         print call_user_func( $fn, 6, 2 );      } }  $b = new Bar; $b->UseReferences();